# Integration partial fractions

• Sep 10th 2011, 03:32 AM
celtic1234
Integration partial fractions
Hi all,
IS this correct?
thanks
John

[math}\int \frac{x.dx}{x^2+4x+13}[/tex]
• Sep 10th 2011, 03:38 AM
e^(i*pi)
Re: Integration partial fractions
Quote:

Originally Posted by celtic1234
Hi all,
IS this correct?
thanks
John

$\displaystyle \int \frac{10}{(2x-6)(x+5)}$

You've only stated the question? Partial fractions is a good method here but first some algebra:

$\int \dfrac{10}{(2x-6)(x+5)} = \int \dfrac{10}{2(x-3)(x+5)} = 5 \int \dfrac{1}{(x-3)(x+5)}$

$\dfrac{1}{(x-3)(x+5)} = \dfrac{A}{x-3} + \dfrac{B}{x+5} \Leftrightarrow 1 = A(x+5) + B(x-3)$

Can you continue?

edit: That other integral is not an easy one to work out. Are you familiar with trig integrals?
• Sep 10th 2011, 03:42 AM
celtic1234
Re: Integration partial fractions
sorry i am busy in the background trying to get laTex to work-i'll try and sort it ASAP
• Sep 10th 2011, 03:43 AM
e^(i*pi)
Re: Integration partial fractions
Quote:

Originally Posted by celtic1234
sorry i am busy in the background trying to get laTex to work-i'll try and sort it ASAP

Your latex is good. You just need to change the tag from $$...$$ to $$...$$
• Sep 10th 2011, 03:45 AM
celtic1234
Re: Integration partial fractions
I am solving the partial fraction to give A= 5/4 and B=-1/6.

The integral solves as:

$\frac{1}{4}5ln(2x-6) - \frac{1}{6}ln(x+1) +c$

Can you tell me if i am correct?
thanks
John
• Sep 10th 2011, 03:55 AM
e^(i*pi)
Re: Integration partial fractions
Quote:

Originally Posted by celtic1234
I am solving the partial fraction to give A= 5/4 and B=-1/6.

The integral solves as:

$\frac{1}{4}5ln(2x-6) - \frac{1}{6}ln(x+1) +c$

Can you tell me if i am correct?
thanks
John

I get B = -1/8

Sub in x=-5: $1 = B(-5-3) = -8B \Leftrightarrow B = -\dfrac{1}{8}$

Sub in x=3: $1 = 8A \Leftrightarrow A = \dfrac{1}{8}$

Hence $5 \int \dfrac{1}{(x-3)(x+5)} = 5 \left(\int \dfrac{1}{8(x-3)} - \int \dfrac{1}{8(x+5)}\right) = \dfrac{5}{8} \left(\int \dfrac{1}{x-3} - \int \dfrac{1}{x+5}\right)$

$=\dfrac{5}{8} \left( \ln(x-3) - \ln(x+5) + C_1\right) = \dfrac{5}{8} \ln \left(\dfrac{x-3}{x+5}\right) + C \text{ where } C = \dfrac{5}{8}C_1$

Perhaps it would be easier to make a new thread since there seem to be two questions on the go here! Alternatively I can delete my posts regarding the original integral
• Sep 10th 2011, 03:58 AM
celtic1234
Re: Integration partial fractions
ok thanks
i'll do a new thread